Posted by Alicia on Thursday, November 18, 2010 at 8:39am.
The last part of the question confuses me, I will solve the question in the traditional way
Let the radius be r m, the height be h m and the volume be V m^3, where V is a constant
We know V = πr^2h
h = V/(πr^2)
Cost = 7.5(top area) + 37.5(bottom area) + 30(side area)
= 7.5πr^2 + 37.5πr^2 + 30(2πr)(h)
= 45πr^2 + 60V/r
d(Cost)/dr = 90πr - 60V/r^2 = 0 for a min of Cost
90πr = 60V/r^2
r^3 = 2(πr^2h)/(3π)
3r = 2h
The ration of radius to height has to be 2 : 3
I will let you deal with the weird condition at the end.
After doing a "search" for your question, I noticed that
MathMate had answered the same question in a much more detailed way.
http://www.jiskha.com/display.cgi?id=1289839685
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